Determine the freezing point and the boiling point of an aqueous solution of LiCl that contains 3.25 g of LiCl and 10.0 mL of water.

Data: K_f [H₂O] = 1.86 K.kg.mol^-1 [freezing point depression constant]

K_b [H₂O] = 0.512 K.kg.mol^-1 [boiling point elevation constant]

Question

Determine the freezing point and the boiling point of an aqueous solution of LiCl that contains 3.25 g of LiCl and 10.0 mL of water.

Data: K_f [H₂O] = 1.86 K.kg.mol^-1 [freezing point depression constant]

K_b [H₂O] = 0.512 K.kg.mol^-1 [boiling point elevation constant]

| General Chemistry 3

Accepted Answer

LiCl (s) → Li⁺ (aq) + Cl^- (aq)

solution molality m = $\frac{n_{solute}}{m_{solution}}$

m = $\frac{m_{LiCl}}{M_{LiCl} \times m_{solution}}$

m = $\frac{3.25}{42.39 \times 10.0 \times 10^{- 3}}$

m = 7.67 mol.kg^-1

The magnitude of the freezing point depression ΔT_f (in K) is:

ΔT_f = - i x K_f x m

ΔT_f = T_f _solution – T_{f water}

i = van’t Hoff i-factor = number of ions produced per formula unit = 2

m = solution molality = 7.67 mol.kg^-1

K_f = proportionality constant (freezing point depression constant)

T_{f solution} – 273 = - 2 x 1.86 x 7.67

T_{f solution} = 244.5 K

The magnitude of the boiling point elevation ΔT_b (in K) is:

ΔT_b = i x K_b x m

ΔT_b = T_{b solution} – T_{b water}

i = van’t Hoff i-factor = number of ions produced per formula unit = 2

m = solution molality (in mol.kg^-1) = 7.67

K_b = proportionality constant (boiling point elevation constant)

T_{f solution} – 373 = 2 x 0.512 x 7.67

T_{f solution} = 380.9 K

Final - Question 8 | General Chemistry 3